Question

rewrite the expression, tan(arcsin x), as an algebraic expression in x

Answer 1

draw a picture of a triangle

Answer 2

marked in an angle whose sine is \(x\) i.e. the angle labelled represents
\[\arcsin(x)\]
you want the tangent of that angle, so you need the unmarked side which you find by pythagoras

Answer 3

the adjacent side is \(\sqrt{1-x^2}\) by pythagoras
you want the tangent of that angle, which is "opposite over adjacent" so you can write
\[\tan(\arctan(x))=\frac{x}{\sqrt{1-x^2}}\]

Answer 4

ok let me get it right
\[\tan(\arcsin(x))=\frac{x}{\sqrt{1-x^2}}\]